This document discusses the dining philosophers problem and solutions using semaphores to avoid deadlock. It describes philosophers taking turns eating with alternating groups being able to eat simultaneously if adjacent philosophers eat one by one. It also explains how a deadlock could occur if all philosophers try to eat at the same time by acquiring forks, with the right forks all being held by other philosophers. Semaphores are initialized to represent each fork, with philosophers following a pattern to acquire the necessary semaphores representing forks without causing a deadlock.

1. Void Philosopher(Void)
{
While(true)
{
Thinking(),
Take-spoon(i),
Take-spoon((i+1)%N),
EAT(),
Put-spoon(i),
Put spoon((i+1)%N),
}
}
p1
p2
p0
p3
p4
code that must be followed by philosopers
f0
f1
f2
f3
f4

2. Changes in function of code while accesing
food by philosophers…..
• Void Philosopher(Void)
• {
• While(true)
• {
• Thinking(),
• Wait(Take-spoon(i)),
• Wait(Take-spoon((i+1)%N)),
• EAT(),
• Signal(Put-spoon(i)),
• Signal(Put spoon((i+1)%N)),
• }
• }

3. Here to solve dp problem ,we will use N no of
Binary Semaphores..
• S[i]
• N=no of spoons
• So here we will use 5 semaphores
• all the semaphores are initialized with 1
•S0 , S1 , S2 , S3 , S4
1 1 1 1 1

4. Philosopher need what semaphore as value 1
at which time?
philosopher 1st clock cycle semaphore 2nd clock cycle semaphore
p0 s0 s1
p1 s1 s2
p2 s2 s3
p3 s3 s4
p4 s4 s0

5. Exception of MUTUAL EXCLUSION
MUTUAL EXCLUSION-
there can be only one process access the critical
section at a time.
BUT , in the dining philosophers problem more than one philosopher can acces
the critical section if they are independent to each other.
For example:-
p0,p2 p1,p3 p2,p4 p5,p1
These pair can eat at same time.

6. 1st Condition:-
If all the adjacent philosopher are about to eat
one after one ,then the alternative group can eat
together,
After that remaining philosopher can eat.

7. 2nd Condition:-
• All philosopher are willing to have food together aat same time.
• And they all have to execute code of entry.
• At the time of 1st clock cycle they will get their left fork f(i)
successfully.
• But at time of 2nd clock cycle when philosophers will try to have
ther right side placed spoon, they all(philosopher will blocked) ,
because all spoons are consumed by previous philosopher.
• So here DEADLOCK occurs .